CoderForces 510B (dfs)

最新推荐文章于 2019-08-17 16:25:00 发布

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最新推荐文章于 2019-08-17 16:25:00 发布

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分类专栏： DFS && BFS

本文链接：https://blog.youkuaiyun.com/qq_34681949/article/details/52204486

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本文介绍了一种算法，用于检测游戏TwoDots中相同颜色的圆点是否构成循环。通过深度优先搜索（DFS）遍历棋盘上的每个圆点，确保找到至少四个相邻且颜色相同的圆点形成的有效循环。

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E - E

Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 510B

Description

Fox Ciel is playing a mobile puzzle game called "Two Dots". The basic levels are played on a board of size n × m cells, like this:

$\text{[math]}$

Each cell contains a dot that has some color. We will use different uppercase Latin characters to express different colors.

The key of this game is to find a cycle that contain dots of same color. Consider 4 blue dots on the picture forming a circle as an example. Formally, we call a sequence of dots d₁, d₂, ..., d_k a cycle if and only if it meets the following condition:

These k dots are different: if i ≠ j then d_i is different from d_j.
k is at least 4.
All dots belong to the same color.
For all 1 ≤ i ≤ k - 1: d_i and d_i + 1 are adjacent. Also, d_k and d₁ should also be adjacent. Cells x and y are called adjacent if they share an edge.

Determine if there exists a cycle on the field.

Input

The first line contains two integers n and m (2 ≤ n, m ≤ 50): the number of rows and columns of the board.

Then n lines follow, each line contains a string consisting of m characters, expressing colors of dots in each line. Each character is an uppercase Latin letter.

Output

Output "Yes" if there exists a cycle, and "No" otherwise.

Sample Input

Input

3 4
AAAA
ABCA
AAAA

Output

Yes

Input

3 4
AAAA
ABCA
AADA

Output

No

Input

4 4
YYYR
BYBY
BBBY
BBBY

Output

Yes

Input

7 6
AAAAAB
ABBBAB
ABAAAB
ABABBB
ABAAAB
ABBBAB
AAAAAB

Output

Yes

Input

2 13
ABCDEFGHIJKLM
NOPQRSTUVWXYZ

Output

No

Hint

In first sample test all 'A' form a cycle.

In second sample there is no such cycle.

The third sample is displayed on the picture above ('Y' = Yellow, 'B' = Blue, 'R' = Red).

代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
char map[51][51];
int vis[51][51];
int dx[4]={0,0,-1,1};
int dy[4]={1,-1,0,0};
char c;
int st,en,n,m;
bool flag;
void dfs(int x,int y,int fx,int fy,char c)//fx,fy用来记录上一个能走的位置，避免向左走的时候重复； 
{
	if(flag==1)
	return ;
	for(int i=0;i<4;i++)
	{
		int a=x+dx[i];
		int b=y+dy[i];
		if(a>=0&&a<n&&b>=0&&b<m&&map[a][b]==c)
		{
			if(a==fx&&b==fy)
			{
				continue;
			}
			if(vis[a][b]==1)//走到起点才成立； 
			{
				flag=1;
				return;
			}
			vis[a][b]=1;//走到一个能走的位置先标记再搜索； 
			dfs(a,b,x,y,map[a][b]);
		}
	}
}
int main()
{
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		for(int i=0;i<n;i++)
		{
			scanf("%s",map[i]);
		}
		memset(vis,0,sizeof(vis));
		flag=0;
		for(int i=0;i<n;i++)
		{
			for(int j=0;j<m;j++)
			{ 
			     int fx=-1,fy=-1; 
			     if(vis[i][j]==0)
			     {
			     	vis[i][j]=1;
			     	dfs(i,j,fx,fy,map[i][j]);
			     	if(flag==1)
			     	break;
				 }
		    }
		    if(flag==1)
		    break;
		}
		if(flag==1)
		printf("Yes\n");
		else
		printf("No\n");
	}
	return 0;
}